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Proof step 8 from proof mod-div-zero (previous step)

0 ≤ j + abs(i) % j, j < 0 ⊦ j = 0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 9:
⊦ j < 0 → abs(j) = 0 - j
Proof step 10:
¬ i < 0, 0 ≤ j + abs(i) % j, j < 0 ⊦ j = 0

Rule argument

spec: int-abs inst: name: Abs-neg seq: ⊦ j < 0 → abs(j) = 0 - j subst: [j] → [i] flags: rewrite

Additional information

Used simplifier rules: mod-neg-01: j ≠ 0 ⊦ ~ i % j = ~(i % j);%mod-not-less: ⊦ j < 0 ∧ i < 0 → ¬ i % j ≤ j;%neg-06: ⊦ 0 - i = ~ i; from specification int-abs%neg-15: ⊦ i + ~ j = i - j; from specification int-abs%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2%sub-le-01: ⊦ i ≤ j - i0 ↔ i + i0 ≤ j; from specification int-abs%add-zero-rev: ⊦ 0 + i = i; from specification int-basic2 Used terms: i